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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Mathematics
Chapter
Sequences and Series

BITSAT 2014 · Mathematics · Sequences and Series

If the (2 p)^ th term of a H . P . is q and the (2 q)^ th term is...

If the (2p)th (2 \mathrm{p})^{\text{th }} term of a H\mathrm{H}. P\mathrm{P}. is q\mathrm{q} and the (2q)th (2 \mathrm{q})^{\text{th }} term is p,\mathrm{p}, then the 2(p+q)th2(\mathrm{p}+\mathrm{q})^{\mathrm{th}} term is-

Options

  1. Apq2(p+q)\frac{p q}{2(p+q)}
  2. B2pqp+q\frac{2 \mathrm{pq}}{\mathrm{p}+\mathrm{q}}
  3. Cpqp+q\frac{\mathrm{pq}}{\mathrm{p}+\mathrm{q}}
  4. Dp+qpq\frac{\mathrm{p}+\mathrm{q}}{\mathrm{pq}}

Correct answer

D. p+qpq\frac{\mathrm{p}+\mathrm{q}}{\mathrm{pq}}

Step-by-step solution

If a is the first term and d is the common difference of the associated A.P. 1q=1a+(2p−1)d,1p=1a+(2q−1)d⇒d=12pq\frac{1}{\mathrm{q}}=\frac{1}{\mathrm{a}}+(2 \mathrm{p}-1) \mathrm{d}, \frac{1}{\mathrm{p}}=\frac{1}{\mathrm{a}}+(2 \mathrm{q}-1) \mathrm{d}\Rightarrow \mathrm{d}=\frac{1}{2 \mathrm{pq}} If h is the 2(p+q)th2(\mathrm{p}+\mathrm{q})^{\mathrm{th}} term 1h=1a+(2p+2q−1)d=1q+1p=p+qpq\frac{1}{\mathrm{ h}}=\frac{1}{\mathrm{a}}+(2 \mathrm{p}+2 \mathrm{q}-1) \mathrm{d}=\frac{1}{\mathrm{q}}+\frac{1}{\mathrm{p}}=\frac{\mathrm{p}+\mathrm{q}}{\mathrm{pq}}
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